Week 5 notes

Week 5 Differentiating $\ln,$ $x^{b},$ $\sin$ and $\cos$

Version 2025/02/26 Week 5 in PDF All notes in PDF To other weeks

We start this chapter by differentiating $\ln (x),$ the natural logarithm function. Since we introduced $\ln$ as the inverse function of $e^{x},$ we will need the following result.

The Inverse Rule of differentiation

Theorem 5.1: The Inverse Rule.

Let $f (x)$ be strictly monotonic and continuous on $[a, b],$ and let $g$ be the inverse of $f$ so that $g$ is strictly monotonic and continuous by the Inverse Function Theorem.

Suppose $f$ is differentiable at $ℓ \in (a, b)$ and that the derivative $f^{'} (ℓ)$ is not zero.

Then $g$ is differentiable at $k = f (ℓ),$ and $g^{'} (k) = \frac{1}{f^{'} (ℓ)} .$

Proof. We begin in the same way as in the proof of the Chain Rule: using Proposition 4.6, write $f (x) - f (ℓ) = F_{ℓ} (x) (x - ℓ)$ for all $x$ and in particular for $x = g (y),$ so

$f (g (y)) - f (g (k)) = F_{ℓ} (g (y)) (g (y) - g (k)) .$

Since $f$ and $g$ are inverse to each other, $f (g (y)) = y,$ so

$y - k = F_{ℓ} (g (y)) (g (y) - g (k)) .$

When $y \neq k,$ we have $y - k \neq 0,$ and the equation shows that $F_{ℓ} (g (y)) \neq 0$ (because the left-hand side is not $0$ ). When $y = k,$ we have $F_{ℓ} (g (y)) = F_{ℓ} (ℓ) = f^{'} (ℓ)$ which is not $0$ by assumption. Hence $F_{ℓ} (g (y))$ is never zero, and we can divide by it:

$\frac{1}{F_{ℓ} (g (y))} (y - k) = g (y) - g (k) .$

Since $\frac{1}{F_{ℓ} (g (y))}$ is continuous at $y = k$ by Algebra of Continuous Functions and continuity of composition, by Proposition 4.6 $g (y)$ is differentiable at $k$ with

$\frac{1}{F_{ℓ} (g (k))} = \frac{1}{F_{ℓ} (ℓ)} = \frac{1}{f^{'} (ℓ)}$

as derivative. □

We immediately deduce

Corollary 5.2: inverse of a function differentiable on an interval.

If $f$ is strictly monotone and differentiable on an open interval, its inverse function $f^{- 1}$ is differentiable everywhere it is defined except the points $f (ℓ)$ with $f^{'} (ℓ) = 0,$ and

$\frac{d (f^{- 1})}{d y} (y) = \frac{1}{\frac{d f}{d x} (f^{- 1} (y))} . \qed$

Can the derivative of a strictly increasing function be zero at some points? Yes:

Example: a point where the inverse to a monotone function is not differentiable.

Construct a continuous strictly increasing function $f (x)$ such that $f^{'} (ℓ) = 0$ for some point $ℓ .$ Check that $f^{- 1}$ is not differentiable at $f (ℓ)$ in your example.

Solution: for example, $f (x) = x^{3}$ and $ℓ = 0.$ We have $f^{'} (0) = 0.$ For the inverse function $\sqrt[3]{y},$ the limit $lim_{y \to 0} \frac{\sqrt[3]{y} - \sqrt[3]{0}}{y - 0} = lim_{y \to 0} \frac{1}{(\sqrt[3]{y})^{2}}$ is $\infty .$ This suggests “infinite derivative” at $0,$ yet formally means that $\sqrt[3]{y}$ is not differentiable at $0.$ Figure 5.1 shows horizontal tangent to the graph of $x^{3},$ and vertical tangent for $\sqrt[3]{x},$ at $(0, 0) .$

Corollary: derivative of $\ln$ .

$\frac{d}{d y} \ln (y) = \frac{1}{y}$ for all $y \in (0, + \infty) .$

Indeed, $\ln$ is the inverse function to $\exp,$ so, using the Inverse Rule and the Theorem which says that $\exp^{'} = \exp,$ we calculate

$\ln^{'} (y) = \frac{1}{\exp^{'} (\ln (y))} = \frac{1}{\exp (\ln (y))} = \frac{1}{y} .$

Functions $x^{b}$ and $a^{x}$

Definition: $a^{b}$ .

For positive real $a$ and real $b$ we define

$a^{b} = e^{b \ln a} .$

Remark: this satisfies the exponent laws $a^{b + c} = a^{b} a^{c},$ $(a c)^{b} = a^{b} c^{b},$ $a^{0} = 1$ and $a^{- b} = 1 / a^{b}$ (exercise: deduce these from the properties of the functions $\ln$ and $\exp$ ). In particular, $a^{n}$ defined as $a \cdot a \cdot \dots \cdot a$ ( $n$ factors) coincides with $e^{n \ln a} .$

Example: derivative of $x^{b}$ .

Fix $b \in R .$ The function $x^{b}$ is defined for positive $x .$ Show that

$\frac{d}{d x} (x^{b}) = b x^{b - 1} .$

Solution. Indeed, by Chain Rule $\frac{d}{d x} (x^{b}) = \frac{d}{d x} (\exp (b \ln (x)) = \exp^{'} (b \ln (x)) \cdot b \ln^{'} (x) .$ Using the derivatives of $\exp (x)$ and of $\ln (x)$ obtained earlier, this is $\exp (b \ln (x)) \cdot b x^{- 1} = x^{b} \cdot b x^{- 1} = b x^{b - 1}$ where we use the exponent laws. □

Exercise. Let $a > 0.$ Use the Chain Rule to show that $\frac{d}{d x} (a^{x}) = a^{x} \ln a .$

The limit definition of $e^{x}$ (optional material)

The section “The limit definition of $e^{x}$ ” was not covered in class and is not examinable.

Although in this course we define $e^{x}$ as the sum $\sum_{k \geq 0} x^{k} / k!,$ originally it was defined via the following limit.

Proposition 5.3: the limit definition for $e^{x}$ (not examinable).

For all $x \in R,$ $e^{x} = lim_{n \to \infty} (1 + \frac{x}{n})^{n} .$ In particular, $e = lim_{n \to \infty} (1 + \frac{1}{n})^{n} .$

Proof. The function $\ln (1 + y)$ is differentiable (by Chain Rule), and its derivative at $y = 0$ is $\frac{1}{1 + 0} = 1.$ By Proposition 4.6 we can write $\ln (1 + y) = F_{0} (y) y$ where the slope function $F_{0}$ is continuous at $0$ with $F_{0} (0) = 1.$ Then for a fixed $x \in R,$

$(1 + \frac{x}{n})^{n} = e^{n \ln (1 + \frac{x}{n})} = e^{n F_{0} (\frac{x}{n}) \frac{x}{n}} = e^{F_{0} (\frac{x}{n}) x} .$

The sequence $y_{n} = \frac{x}{n}$ is monotone with limit $0$ so Lemma 1.1 and its Corollary tell us that $lim_{n \to \infty} e^{F_{0} (y_{n}) x}$ is $lim_{y \to 0} e^{F_{0} (y) x} .$ Since $e^{F_{0} (y) x}$ is a continuous function of $y,$ by criterion of continuity its limit as $y \to 0$ equals its value at $0,$ which is $e^{F_{0} (0) x} = e^{1 \cdot x} = e^{x} .$ □

End of the non-examinable section.

The functions $\sin$ and $\cos$

In this course, $\sin 𝛼$ and $\cos 𝛼$ are defined as the ratio of sides in a right-angled triangle. Thus, if $P_{𝛼}$ is the point on the unit circle centred at the origin $O$ such that the angle between the $x$ axis and $O P_{𝛼}$ is $𝛼,$ then $P_{𝛼}$ has coordinates $(\cos 𝛼, \sin 𝛼) .$ This defines the sine and cosine of all $𝛼 \in R,$ see Fig. 5.2. Note that $𝛼$ is in radians: the angle equal to the full revolution (full circle) is $2 𝜋 .$ The definition implies that for all $𝛼 \in R,$

$\cos (- 𝛼) = \cos 𝛼, \sin (- 𝛼) = - \sin 𝛼, \cos 𝛼 = \sin (\frac{𝜋}{2} - 𝛼) .$

We also define $\tan 𝛼 = \frac{\sin 𝛼}{\cos 𝛼}$ whenever $\cos 𝛼 \neq 0.$ When $𝛼 \in (- \frac{𝜋}{2}, \frac{𝜋}{2}),$ the straight line $O P_{𝛼}$ intersects the tangent to the circle at $P_{0}$ at the point $T_{𝛼} (1, \tan 𝛼),$ see Fig. 5.2.

Lemma 5.4: the sine-angle-tangent inequality.

$0 \leq \sin 𝛼 \leq 𝛼 \leq \tan 𝛼$ for all $𝛼 \in (0, \frac{𝜋}{2}) .$

Proof. In Fig. 5.2, area( $△ O P_{0} P_{𝛼}$ ) $=$ $\frac{1}{2}$ $\times$ base $\times$ height $= \frac{1}{2} \times 1 \times \sin 𝛼 = \frac{1}{2} \sin 𝛼 .$

The area of the sector $O P_{0} P_{𝛼}$ with central angle $𝛼$ is $\frac{𝛼}{2 𝜋}$ $\times$ area(circle) $=$ $\frac{𝛼}{2 𝜋} \times 𝜋 = \frac{1}{2} 𝛼 .$

Area( $△ O P_{0} T_{𝛼}$ ) is $\frac{1}{2} \times 1 \times \tan 𝛼 .$ We have $△ O P_{0} P_{𝛼}$ $\subseteq$ sector $O P_{0} P_{𝛼}$ $\subseteq$ $△ O P_{0} T_{𝛼},$ therefore $\frac{1}{2} \sin 𝛼 \leq \frac{1}{2} 𝛼 \leq \frac{1}{2} \tan 𝛼 .$ □

Remark. Graphs in Fig. 5.3 illustrate the behaviour of $\sin x,$ $x$ and $\tan x$ for small $x .$

Corollary 5.5: limit of sine at $0$ .

$lim_{x \to 0} | \sin x | = 0.$

Proof.

The inequality in Lemma 5.4 can be written as $0 \leq | \sin x | \leq x$ when $x \in (0, \frac{𝜋}{2}),$ as $\sin x$ is positive for these $x .$ Since $lim_{x \to 0 +} 0 = lim_{x \to 0 +} x = 0,$ we have $lim_{x \to 0 +} | \sin x | = 0$ by Sandwich Rule.

Putting $t = - x,$ we have $lim_{x \to 0 -} | \sin x | = lim_{t \to 0 +} | \sin (- t) | = lim_{t \to 0 +} | \sin t | = 0.$

The one-sided limits exist and are equal, so $lim_{x \to 0} | \sin x |$ is their common value $0.$ □

We need another result about $\sin$ and $\cos$ with a geometric proof.

Lemma 5.6: sine and cosine subtraction formulas.

i. $\sin y - \sin x = 2 \sin (\frac{y - x}{2}) \cos (\frac{y + x}{2});$
ii. $\cos y - \cos x = - 2 \sin (\frac{y - x}{2}) \sin (\frac{y + x}{2}) .$

Proof of Lemma 5.6 — proof not examinable. If $M$ is the midpoint of the segment $P_{𝛼} P_{𝛽},$ the vector $\vec{O M}$ is $\frac{1}{2} ({\vec{O P}}_{𝛼} + {\vec{O P}}_{𝛽})$ and so the $y$ -coordinate of $M$ is $\frac{1}{2} (\sin 𝛼 + \sin 𝛽) .$ On the other hand, see Fig. 5.4, $\vec{O M}$ is proportional to ${\vec{O P}}_{(𝛼 + 𝛽) / 2}$ and, from the right-angled triangle $△ O M P_{𝛽}$ we can see that $\vec{O M} = \cos (\frac{𝛼 - 𝛽}{2}) {\vec{O P}}_{(𝛼 + 𝛽) / 2} .$ This expresses the $y$ -coordinate of $M$ via the $y$ -coordinate of $P_{(𝛼 + 𝛽) / 2} :$

$\frac{1}{2} (\sin 𝛼 + \sin 𝛽) = \cos (\frac{𝛼 - 𝛽}{2}) \sin (\frac{𝛼 + 𝛽}{2}) .$

This sine addition formula holds for all $𝛼, 𝛽 \in R .$ Substitute $𝛼 = y,$ $𝛽 = - x$ to obtain the sine subtraction formula i. as claimed.

Substitute $𝛼 = \frac{𝜋}{2} - y,$ $𝛽 = x - \frac{𝜋}{2} :$ the LHS becomes $\frac{1}{2} (\sin (\frac{𝜋}{2} - y) - \sin (\frac{𝜋}{2} - x))$ which is $\frac{1}{2} (\cos y - \cos x),$ the RHS becomes $\cos (\frac{𝜋}{2} - \frac{y + x}{2}) \sin (\frac{- y + x}{2}) = - \sin (\frac{y + x}{2}) \sin (\frac{y - x}{2}),$ equivalent to the cosine subtraction formula ii. □

Proposition 5.7: continuity of $\sin$ and $\cos$ .

$\sin x$ and $\cos x$ are continuous functions on $R .$

Proof. In the sine subtraction formula (Lemma 5.6), we bound the modulus of $\cos$ by $1 :$

$| \sin y - \sin x | = 2 | \sin \frac{y - x}{2} | | \cos \frac{y + x}{2} | \leq 2 | \sin h |,$

where we put $h = (y - x) / 2.$ Opening out the modulus, we get

$- 2 | \sin h | \leq \sin y - \sin x \leq 2 | \sin h | .$

Taking the limit as $y \to x,$ equivalently $h \to 0,$ we have $lim_{h \to 0} 2 | \sin h | = 0$ by Corollary 5.5. Hence by Sandwich Rule

$lim_{y \to x} \sin (y) - \sin (x) = 0 \Leftrightarrow lim_{y \to x} \sin (y) = \sin (x),$

so by the criterion of continuity, $\sin$ is continuous at $x .$

Continuity of $\cos$ is proved similarly and is left to the student. □

Comparing the graphs of $\sin x$ and $x$ in Fig. 5.3, we suspect that these two functions have the same gradient at $0.$ Let us formalise this.

Proposition 5.8: special limit for sine.

$lim_{x \to 0} \frac{\sin x}{x} = 1.$

Proof. We compute the one-sided limits. If $x \in (0, \frac{𝜋}{2}),$ we have

$\cos x = \frac{\sin x}{\tan x} \leq \frac{\sin x}{x} \leq \frac{\sin x}{\sin x} = 1$

from the sine-angle-tangent inequality in Lemma 5.4. As $x \to 0 +,$ $\cos x \to \cos 0 = 1$ because $\cos$ is a continuous function. Hence by Sandwich Rule $lim_{x \to 0 +} \frac{\sin x}{x} = 1.$

In $lim_{x \to 0 -} \frac{\sin x}{x},$ change the variable, $x = - y .$ The limit becomes $lim_{y \to 0 +} \frac{\sin (- y)}{- y} = lim_{y \to 0 +} \frac{\sin y}{y} = 1.$

Both one-sided limits are equal to $1,$ so $lim_{x \to 0} \frac{\sin x}{x}$ exists and equals $1,$ as claimed. □

Theorem 5.9: differentiation of $\sin$ and $\cos$ .

Functions $\sin x$ and $\cos x$ are differentiable everywhere on $R,$ and

$\sin^{'} x = \cos x, \cos^{'} x = - \sin x .$

Proof. Use the sine subtraction formula to write $\sin y - \sin x = 2 \sin h \cos (x + h)$ with $h = (y - x) / 2.$ Then

$\frac{\sin y - \sin x}{y - x} = \frac{2 \sin h \cos (x + h)}{2 h} = \frac{\sin h}{h} \cos (x + h),$

and so

$\sin^{'} x = lim_{h \to 0} \frac{\sin h}{h} \times lim_{h \to 0} \cos (x + h) = 1 \times \cos (x + 0) = \cos x$

by AoL, the Special Limit for sine and continuity of $\cos .$

To differentiate $\cos,$ use $\cos y - \cos x = - 2 \sin h \sin (x + h)$ and conclude similarly. □

Version 2025/02/26 Week 5 in PDF All notes in PDF To other weeks

Week 5 Differentiating ln, xb, sin and cos

The Inverse Rule of differentiation

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The limit definition of ex (optional material)

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Week 5 Differentiating $\ln,$ $x^{b},$ $\sin$ and $\cos$

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The limit definition of $e^{x}$ (optional material)

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